the discharge time of 10uF capacitor which put in the VDD pin of 51822

[jessie]

Hi all;

I have a hardware question . How much is the discharge time of 10uF capacitor which put in the VDD pin of 51822 ?

When i use the formula of capacitor discharge time , it is not need much time .

I put the ammeter in series with my product and apply the power for them, at this time , the 51822 is initialized and 4mA current consumption about 1S, and then power off mode. But after i remove the power about 10s, i reapply the power, 51822 is not initialized , it just only in power off mode. if i remove the power more than 30S , and reapply the power , 51822 is initialized. When it is in power off mode ,the current is about 0.33uA.

So i can not understand that why the discharge time of 10uF capacitor is so long? if anyone was the same as me? Thanks. Who can give me a correct way ?

[bjorn-spockeli]

The voltage across a capacitor with constant current is given by

U(t) = U(t=0) + (I*t)/C ) 

Solving the equation for t to obtain the discharge time

t = C*( U- U(t=0) )/I

The power management system of the chip will put the chip in a reset state if the supply voltage is below 1.7 V, i.e. Brown-out reset.

Thus with U = 1.7V, U(t=0) = 3V, I = -0.33uA, C = 10uF the discharge time is

t = 10uF * (1.7V -3.0V)/-0.33uA = 39s

and the voltage after 10 seconds will be

U(t=10) = 3V + (-0.33uA*10s)/10uF = 2.67 V

[jessie]

Clearly. Thank you very much.